subroutine

Hello every one,
I want to obtain the fortran code of an equation as subroutine.Please
can compile the following equation¨:
τ=m k√[1-e^(-(μ σ /m k)^n ) ]
Where,
Ï„ : frictional stress
σ : noramal stress
m : friction factor, 0<m<1
μ : friction coefficient
k : shear flow stress
n : friction law exponent, ranging between 1 and about 3
Thank you very much for your time,
Best regards,
Mohamed

On 4/28/2012 12:19 PM, yahoo.com">eng_mohamedramadan@yahoo.com wrote:
> Hello every one,
> I want to obtain the fortran code of an equation as subroutine.Please
> can compile the following equation¨:
> τ=m k√[1-e^(-(μ σ /m k)^n ) ]
....

Seems too trivial for a posting; what's the _real_ question?

Choose names for the variables (remember to declare them) of the
appropriate type.

Pass them as arguments to let the routine receive specific values for
each invocation.

Multiplication is "*", exponentiation is "**" and division is "/"
The intrinsics for square root and the exponential are SQRT and EXP,
respectively.

Grouping of operations to get the proper associativity is simply
multiple parentheses as written other than the square brackets will have
to also be parens.

Have a go at it; bound to be good practice if are just beginning...

--

On Apr 28, 7:01Â*pm, dpb <n...@non.net> wrote:
> On 4/28/2012 12:19 PM, eng_mohamedrama...@yahoo.com wrote:> Hello every one,
> > I want to obtain the fortran code of an equation as subroutine.Please
> > can compile the following equation¨:
> > Ï„=m k√[1-e^(-(μ σ /m Â*k)^n ) ]
>
> ...
>
> Seems too trivial for a posting; what's the _real_ question?
>
> Choose names for the variables (remember to declare them) of the
> appropriate type.
>
> Pass them as arguments to let the routine receive specific values for
> each invocation.
>
> Multiplication is "*", exponentiation is "**" and division is "/"
> The intrinsics for square root and the exponential are SQRT and EXP,
> respectively.
>
> Grouping of operations to get the proper associativity is simply
> multiple parentheses as written other than the square brackets will have
> to also be parens.
>
> Have a go at it; bound to be good practice if are just beginning...
>
> --

Thank you, yes according to your words, it is seem trivial. Sorry, but
I dont study fortran before.

On 4/28/2012 1:14 PM, yahoo.com">eng_mohamedramadan@yahoo.com wrote:
....

> Thank you, yes according to your words, it is seem trivial. Sorry, but
> I dont study fortran before.

Well, no time like the present...

<http://francesa.phy.cmich.edu/people/fornari/pdf/f90tutor.pdf>

--

On 4/28/12 1:14 PM, yahoo.com">eng_mohamedramadan@yahoo.com wrote:
> On Apr 28, 7:01 pm, dpb<n...@non.net> wrote:
>> On 4/28/2012 12:19 PM, eng_mohamedrama...@yahoo.com wrote:> Hello every one,
>>> I want to obtain the fortran code of an equation as subroutine.Please
>>> can compile the following equation¨:
>>> τ=m k√[1-e^(-(μ σ /m k)^n ) ]
>>
>> ...
>>
>> Seems too trivial for a posting; what's the _real_ question?
>>
>> Choose names for the variables (remember to declare them) of the
>> appropriate type.
>>
>> Pass them as arguments to let the routine receive specific values for
>> each invocation.
>>
>> Multiplication is "*", exponentiation is "**" and division is "/"
>> The intrinsics for square root and the exponential are SQRT and EXP,
>> respectively.
>>
>> Grouping of operations to get the proper associativity is simply
>> multiple parentheses as written other than the square brackets will have
>> to also be parens.
>>
>> Have a go at it; bound to be good practice if are just beginning...
>>
>> --
>
> Thank you, yes according to your words, it is seem trivial. Sorry, but
> I dont study fortran before.
OK, why don't you try something and post it here, somebody will be
willing to help.

As a hint, Fortran is pretty tricky with parenthesis and operator
ordering (precedence) in expressions.

There's a big difference between
-( (Sigma*Mu)/(M*K) )**N and (I think this is what you want)
-( (Sigma*Mu)/ M*K )**N and
- (Sigma*Mu)/(M*K) **N

Dick Hendrickson

On 4/28/2012 2:45 PM, Dick Hendrickson wrote:
> On 4/28/12 1:14 PM, yahoo.com">eng_mohamedramadan@yahoo.com wrote:
>> On Apr 28, 7:01 pm, dpb<n...@non.net> wrote:
>>> On 4/28/2012 12:19 PM, eng_mohamedrama...@yahoo.com wrote:> Hello
>>> every one,
>>>> I want to obtain the fortran code of an equation as subroutine.Please
>>>> can compile the following equation¨:
>>>> τ=m k√[1-e^(-(μ σ /m k)^n ) ]
>>>
>>> ...
>>>
>>> Seems too trivial for a posting; what's the _real_ question?
>>>
>>> Choose names for the variables (remember to declare them) of the
>>> appropriate type.
>>>
>>> Pass them as arguments to let the routine receive specific values for
>>> each invocation.
>>>
>>> Multiplication is "*", exponentiation is "**" and division is "/"
>>> The intrinsics for square root and the exponential are SQRT and EXP,
>>> respectively.
>>>
>>> Grouping of operations to get the proper associativity is simply
>>> multiple parentheses as written other than the square brackets will have
>>> to also be parens.
>>>
>>> Have a go at it; bound to be good practice if are just beginning...
>>>
>>> --
>>
>> Thank you, yes according to your words, it is seem trivial. Sorry, but
>> I dont study fortran before.
> OK, why don't you try something and post it here, somebody will be
> willing to help.
>
> As a hint, Fortran is pretty tricky with parenthesis and operator
> ordering (precedence) in expressions.
>
> There's a big difference between
> -( (Sigma*Mu)/(M*K) )**N and (I think this is what you want)
> -( (Sigma*Mu)/ M*K )**N and
> - (Sigma*Mu)/(M*K) **N
>
> Dick Hendrickson
Why tricky? Isn't it pretty much standard mathematics?

On Apr 28, 9:53Â*pm, Gary Scott <garylsc...@sbcglobal.net> wrote:
> On 4/28/2012 2:45 PM, Dick Hendrickson wrote:
>
>
>
>
>
>
>
> > On 4/28/12 1:14 PM, eng_mohamedrama...@yahoo.com wrote:
> >> On Apr 28, 7:01 pm, dpb<n...@non.net> wrote:
> >>> On 4/28/2012 12:19 PM, eng_mohamedrama...@yahoo.com wrote:> Hello
> >>> every one,
> >>>> I want to obtain the fortran code of an equation as subroutine.Please
> >>>> can compile the following equation¨:
> >>>> τ=m k√[1-e^(-(μ σ /m k)^n ) ]
>
> >>> ...
>
> >>> Seems too trivial for a posting; what's the _real_ question?
>
> >>> Choose names for the variables (remember to declare them) of the
> >>> appropriate type.
>
> >>> Pass them as arguments to let the routine receive specific values for
> >>> each invocation.
>
> >>> Multiplication is "*", exponentiation is "**" and division is "/"
> >>> The intrinsics for square root and the exponential are SQRT and EXP,
> >>> respectively.
>
> >>> Grouping of operations to get the proper associativity is simply
> >>> multiple parentheses as written other than the square brackets will have
> >>> to also be parens.
>
> >>> Have a go at it; bound to be good practice if are just beginning...
>
> >>> --
>
> >> Thank you, yes according to your words, it is seem trivial. Sorry, but
> >> I dont study fortran before.
> > OK, why don't you try something and post it here, somebody will be
> > willing to help.
>
> > As a hint, Fortran is pretty tricky with parenthesis and operator
> > ordering (precedence) in expressions.
>
> > There's a big difference between
> > -( (Sigma*Mu)/(M*K) )**N and (I think this is what you want)
> > -( (Sigma*Mu)/ M*K )**N and
> > - (Sigma*Mu)/(M*K) **N
>
> > Dick Hendrickson
>
> Why tricky? Â*Isn't it pretty much standard mathematics?

I try to do it as following but I find error, can you help me?
SUBROUTINE ADD(Tau,m,k,Sigma,Mu,n)
Tau =m*k*SQRT(1.0-EXP((-Sigma*Mu)/(M*K))**N)
M>0.0.AND.M<1.0
N>=1.0.AND.N<=3.0
RETURN
END

On 4/28/12 3:53 PM, Gary Scott wrote:
> On 4/28/2012 2:45 PM, Dick Hendrickson wrote:
>> On 4/28/12 1:14 PM, yahoo.com">eng_mohamedramadan@yahoo.com wrote:
>>> On Apr 28, 7:01 pm, dpb<n...@non.net> wrote:
>>>> On 4/28/2012 12:19 PM, eng_mohamedrama...@yahoo.com wrote:> Hello
>>>> every one,
>>>>> I want to obtain the fortran code of an equation as subroutine.Please
>>>>> can compile the following equation¨:
>>>>> τ=m k√[1-e^(-(μ σ /m k)^n ) ]
>>>>
>>>> ...
>>>>
>>>> Seems too trivial for a posting; what's the _real_ question?
>>>>
>>>> Choose names for the variables (remember to declare them) of the
>>>> appropriate type.
>>>>
>>>> Pass them as arguments to let the routine receive specific values for
>>>> each invocation.
>>>>
>>>> Multiplication is "*", exponentiation is "**" and division is "/"
>>>> The intrinsics for square root and the exponential are SQRT and EXP,
>>>> respectively.
>>>>
>>>> Grouping of operations to get the proper associativity is simply
>>>> multiple parentheses as written other than the square brackets will
>>>> have
>>>> to also be parens.
>>>>
>>>> Have a go at it; bound to be good practice if are just beginning...
>>>>
>>>> --
>>>
>>> Thank you, yes according to your words, it is seem trivial. Sorry, but
>>> I dont study fortran before.
>> OK, why don't you try something and post it here, somebody will be
>> willing to help.
>>
>> As a hint, Fortran is pretty tricky with parenthesis and operator
>> ordering (precedence) in expressions.
>>
>> There's a big difference between
>> -( (Sigma*Mu)/(M*K) )**N and (I think this is what you want)
>> -( (Sigma*Mu)/ M*K )**N and
>> - (Sigma*Mu)/(M*K) **N
>>
>> Dick Hendrickson
> Why tricky? Isn't it pretty much standard mathematics?

OK, maybe tricky is too strong. But, the OP is a beginner and in his
expression
(μ σ /m k)
you can't translate that without being careful with "k"

Dick Hendrickson

> > On 4/28/12 1:14 PM, eng_mohamedrama...@yahoo.com wrote:

I try to do it as following but I find error, can you help me?
SUBROUTINE ADD(Tau,m,k,Sigma,Mu,n)
Tau =m*k*SQRT(1.0-EXP((-Sigma*Mu)/(M*K))**N)
M>0.0.AND.M<1.0
N>=1.0.AND.N<=3.0
RETURN
END

Now you have a new problem.

But first, YOU MUST READ A MANUAL on Fortran programming!
The subroutine you wrote before, and this new version, show that you do not
understand the first elements of programming in a mathematical language.

You can start with this subroutine that you have shown us, as an expression
of the ideas you have about how to solve the physical or mathematical
problem you are facing.
Then having written the conditions and basic formula connecting the known
variables (those on the right of the equals sign) to obtain the unknown
variable (Tau), seen on the left of the equals sign, you can proceed to turn
this into a problem-solving program..

But you still have to obtain, or provide, the values of the variables m,k,
Sigma,Mu, and N.
Not that 'm' is the same as 'M' and 'k' is the same as 'K' since Fortran
does not take any notice of the case of the names of the variables.
Also, you have noted that some variables (M.N) are only known to be within
certain ranges, (or else you want to explore the resulting values of Tau
within this two-dimensional domain of M,N as they are varied.

Now, find a book of Fortran programming (it can be a very old one on
Fortran77, or later F90 or F95), but everything you need will be there, and
any Fortran compiler of today can compile and run code written within the
simple limitations you will need.